STD 12 - 2. inverse trigonometric function — JEE कक्षा 11 साइन्स — Question

$ = {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + \cos 2\theta } + \sqrt {1 - \cos 2\theta } }}{{\sqrt {1 + \cos 2\theta } - \sqrt {1 - \cos 2\theta } }}} \right]$

(${x^2} = \cos 2\theta $ रखने पर  $\Rightarrow \theta = \frac{1}{2}{\cos ^{ - 1}}{x^2})$

$= {\tan ^{ - 1}}\left[ {\frac{{\sqrt 2 \cos \theta + \sqrt 2 \sin \theta }}{{\sqrt 2 \cos \theta - \sqrt 2 \sin \theta }}} \right]$

$ = {\tan ^{ - 1}}\left[ {\frac{{1 + \tan \theta }}{{1 - \tan \theta }}} \right] = {\tan ^{ - 1}}\left[ {\frac{{\tan \frac{\pi }{4} + \tan \theta }}{{1 - \tan \frac{\pi }{4}\tan \theta }}} \right]$

$ = {\tan ^{ - 1}}\tan \left( {\frac{\pi }{4} + \theta } \right) = \frac{\pi }{4} + \theta = \frac{\pi }{4} + \frac{1}{2}{\cos ^{ - 1}}{x^2}$.